Parallel and Perpendicular Lines

Joined
Nov 10, 2013
Messages
23
Greetings folks - I'm really struggling with this and our course book is practically useless. The examples tend to be bare bones and somehow we're supposed to figure out the more complicated stuff on our own.

Without further ado:

Find an equation of a line through the given point and (a) parallel to and (b) perpendicular to the given line.

y = 3(x - 2) + 1 at (0, 3)

And my work:

a) y - y1 = 3[(x - 2) - x1]
y - 3 = 3[(x - 2) - 0]
y - 3 = 3x - 6
y = 3x - 3

However, the online calculator I used to check the work says the answer is: y = 3x + 3

Not sure where I might have gone wrong or if the calculator was rubbish.

b) m = -1/3
y - y1 = -1/3[(x - 2) - x1]
y - 3 = -1/3[(x - 2) - 0]
y - 3 = -1/3x + 2/3
y = -1/3x + 3 2/3

Again, I was somewhat close, but the online calculator came up with: y = -1/3x + 3

Thanks for any help, I'm really stressing out over this!
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,258
Greetings folks - I'm really struggling with this and our course book is practically useless. The examples tend to be bare bones and somehow we're supposed to figure out the more complicated stuff on our own.

Without further ado:

Find an equation of a line through the given point and (a) parallel to and (b) perpendicular to the given line.

y = 3(x - 2) + 1 at (0, 3)

And my work:

a) y - y1 = 3[(x - 2) - x1]
y - 3 = 3[(x - 2) - 0]
y - 3 = 3x - 6
y = 3x - 3

However, the online calculator I used to check the work says the answer is: y = 3x + 3

Not sure where I might have gone wrong or if the calculator was rubbish.

b) m = -1/3
y - y1 = -1/3[(x - 2) - x1]
y - 3 = -1/3[(x - 2) - 0]
y - 3 = -1/3x + 2/3
y = -1/3x + 3 2/3

Again, I was somewhat close, but the online calculator came up with: y = -1/3x + 3

Thanks for any help, I'm really stressing out over this!
I am having trouble understanding what you are doing so I am not going to try to correct your work. This is an amazingly easy exercise if you know how to find the slope and y-intercept if an equation is in slope intercept form and you know two facts.

Fact 1: if line A has a slope of m and line B is parallel to line A, then line B also has a slope of m.

First, put the equation of line A into proper slope-intercept form.

\(\displaystyle Line\ A:\ y = 3(x - 2) + 1 = 3x - 6 + 1 = 3x - 5.\) The slope is 3.

\(\displaystyle So\ Line\ B:\ y = 3x + b.\) And b is the y-intercept. So b = what?

Faxt 2: if line A has a slope of m and line C is perpendicular to line A, then line C has a slope of - (1/m).

We already know that the slope of line A is 3. So what is the slope of line C? What is the intercept of line C? So what is the equation of line C
 
Last edited:
Joined
Nov 10, 2013
Messages
23
I am having trouble understanding what you are doing so I am not going to try to correct your work.
I was trying to solve using Point-Slope Form.

This is an amazingly easy exercise if you know how to find the slope and y-intercept if an equation is in slope intercept form and you know two facts.
You're right about that, I see what I did wrong. I was given y = 3(x - 2) + 1, and didn't think to work that out first to y = 3x - 5, and then plug in the points. So here's what I have now, which gives me the same answers as the online calculator:

y = 3(x - 2) + 1
y = 3x - 6 + 1
y = 3x - 5
y - y1 = 3(x - x1)
y - 3 = 3(x - 0)
y - 3 = 3x - 0
y = 3x + 3

m = -1/3
y - y1 = -1/3(x - x1)
y - 3 = -1/3(x - 0)
y - 3 = -1/3x - 0
y = -1/3 + 3


Thanks for your help, I'm sure I'll be back with more questions soon!
 

lookagain

Senior Member
Joined
Aug 22, 2010
Messages
2,396
m = -1/3

y - y1 = (-1/3)(x - x1)

y - 3 = (-1/3)(x - 0)

y - 3 = (-1/3)x - 0

y = -1/3 + 3 \(\displaystyle \ \ \ \ \) You are missing the "x" here.

To be clearer, you should place grouping symbols, such as parentheses, around the fractions as in:


y = (-1/3)x + 3
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,874
I was trying to solve using Point-Slope Form.

You're right about that, I see what I did wrong. I was given y = 3(x - 2) + 1, and didn't think to work that out first to y = 3x - 5, and then plug in the points. So here's what I have now, which gives me the same answers as the online calculator:

y = 3(x - 2) + 1
y = 3x - 6 + 1
y = 3x - 5
I can't imagine why you that this was important. You were given that the point was (0, 3) and you can see immediately from either "y= 3(x- 2)+ 1" or "y= 3x- 5" that the slope is 3.

y - y1 = 3(x - x1)
y - 3 = 3(x - 0)
y - 3 = 3x - 0
y = 3x + 3

m = -1/3
y - y1 = -1/3(x - x1)
y - 3 = -1/3(x - 0)
y - 3 = -1/3x - 0
y = -1/3 + 3


Thanks for your help, I'm sure I'll be back with more questions soon!
 
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